Question: Simplify the following expression: $\dfrac{110k}{33k^2}$ You can assume $k \neq 0$.
Answer: $ \dfrac{110k}{33k^2} = \dfrac{110}{33} \cdot \dfrac{k}{k^2} $ To simplify $\frac{110}{33}$ , find the greatest common factor (GCD) of $110$ and $33$ $110 = 2 \cdot 5 \cdot 11$ $33 = 3 \cdot 11$ $ \mbox{GCD}(110, 33) = 11 $ $ \dfrac{110}{33} \cdot \dfrac{k}{k^2} = \dfrac{11 \cdot 10}{11 \cdot 3} \cdot \dfrac{k}{k^2} $ $\phantom{ \dfrac{110}{33} \cdot \dfrac{1}{2}} = \dfrac{10}{3} \cdot \dfrac{k}{k^2} $ $ \dfrac{k}{k^2} = \dfrac{k}{k \cdot k} = \dfrac{1}{k} $ $ \dfrac{10}{3} \cdot \dfrac{1}{k} = \dfrac{10}{3k} $